By our initial assumptions it follows that \(G^{(n)}\)  is compact since it is closed and is contained inside \(G^{(1)}\times \cdots \times G^{(1)}\) . From [1]} we know that the Lie groupoid \(G^{(1)} \rightrightarrows G^{(0)}\)  admits an \(n\) -metric \(\eta ^{(n)}\)  on \(G^{(n)}\) . It is important to remember that we can push \(\eta ^{(n)}\)  forward with the different face maps \(d_k^n:G^{(n)}\rightarrow G^{(n-1)}\)  to define an \((n-1)\) -metric \(\eta ^{(n-1)}\)  on \(G^{(n-1)}\)  where \(\eta ^{(n-1)}=(d_k^n)_\ast \eta ^{(n)}=(d_{k^{\prime }}^n)_\ast \eta ^{(n)}\)  for all \(k,k^{\prime }\)  and every \(d_k^n:(G^{(n)},\eta ^{(n)})\rightarrow (G^{(n-1)},\eta ^{(n-1)})\)  becomes a Riemannian submersion. One can use this process to obtain \(r\) -metrics \(\eta ^{(r)}\)  on \(G^{(r)}\)  in such a way that \(d_k^r:(G^{(r)},\eta ^{(r)})\rightarrow (G^{(r-1)},\eta ^{(r-1)})\)  is a Riemannian submersion for every \(0\le r\le n-1\) .
